43: Bohr's Model And Physics Of The Atom : Cbse-xi-physics

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CBSE-XI-Physics

43: Bohr's Model and Physics of the Atom

with Solutions - page 2

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Qstn #3
Suppose, the electron in a hydrogen atom makes transition from n = 3 to n = 2 in 10^-8 s. The order of the torque acting on the electron in this period, using the relation between torque and angular momentum as discussed in the chapter on rotational mechanics is
(a) 10^-34 N m
(b) 10^-24 N m
(c) 10^-42 N m
(d) 10^-8 N m
digAnsr: c
Ans : (c) 10^-42N-m
The angular momentum of the electron for the nth state is given by
`` {L}_{\,\mathrm{\,n\,}}=\frac{n\,\mathrm{\,h\,}}{2\,\mathrm{\,\pi \,}}``
Angular momentum of the electron for n = 3, `` {L}_{\,\mathrm{\,i\,}}=\frac{3h}{2\,\mathrm{\,\pi \,}}``
Angular momentum of the electron for n = 2, `` {L}_{\,\mathrm{\,f\,}}=\frac{2h}{2\,\mathrm{\,\pi \,}}``
The torque is the time rate of change of the angular momentum.
`` \,\mathrm{\,Torque\,},\tau =\frac{{L}_{\,\mathrm{\,f\,}}-{L}_{\,\mathrm{\,i\,}}}{t}``
`` =\frac{(2\,\mathrm{\,h\,}/2\,\mathrm{\,\pi \,})-(3\,\mathrm{\,h\,}/2\,\mathrm{\,\pi \,})}{{10}^{-8}}``
`` =\frac{-(\,\mathrm{\,h\,}/2\,\mathrm{\,\pi \,})}{{10}^{-8}}``
`` =\frac{-{10}^{-34}}{{10}^{-8}}\left[\because \frac{\,\mathrm{\,h\,}}{2\,\mathrm{\,\pi \,}}\approx {10}^{-34}\,\mathrm{\,J\,}-\,\mathrm{\,s\,}\right]``
`` =-{10}^{-42}\,\mathrm{\,N\,}-\,\mathrm{\,m\,}``
`` ``
The magnitude of the torque is 10^{`` -``42}N-m.
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Qstn #4
In which of the following transitions will the wavelength be minimum?
(a) n = 5 to n = 4
(b) n = 4 to n = 3
(c) n = 3 to n = 2
(d) n = 2 to n = 1
digAnsr: d
Ans : (d) n = 2 to n = 1
For the transition in the hydrogen-like atom, the wavelength of the emitted radiation is calculated by
`` \frac{1}{\lambda }=\,\mathrm{\,R\,}{Z}^{2}\left(\frac{1}{{n}_{1}}-\frac{1}{{n}_{2}}\right)``
Here, R is the Rydberg constant.
For the transition from n = 5 to n = 4, the wavelength is given by
`` \frac{1}{\lambda }=\,\mathrm{\,R\,}{Z}^{2}\left(\frac{1}{{4}^{2}}-\frac{1}{{5}^{2}}\right)``
`` \lambda =\frac{400}{9\,\mathrm{\,R\,}{Z}^{2}}``
For the transition from n = 4 to n = 3, the wavelength is given by
`` \frac{1}{\lambda }=\,\mathrm{\,R\,}{Z}^{2}\left(\frac{1}{{3}^{2}}-\frac{1}{{4}^{2}}\right)``
`` \lambda =\frac{144}{7\,\mathrm{\,R\,}{Z}^{2}}``
For the transition from n = 3 to n = 2, the wavelength is given by
`` \frac{1}{\lambda }=\,\mathrm{\,R\,}{Z}^{2}\left(\frac{1}{{2}^{2}}-\frac{1}{{3}^{2}}\right)``
`` \lambda =\frac{36}{5\,\mathrm{\,R\,}{Z}^{2}}``
For the transition from n = 2 to n = 1, the wavelength is given by
`` \frac{1}{\lambda }=\,\mathrm{\,R\,}{Z}^{2}\left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)``
`` \lambda =\frac{2}{\,\mathrm{\,R\,}{Z}^{2}}``
From the above calculations, it can be observed that the wavelength of the radiation emitted for the transition from n = 2 to n = 1 will be minimum.
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Qstn #5
In which of the following systems will the radius of the first orbit (n = 1) be minimum?
(a) Hydrogen atom
(b) Deuterium atom
(c) Singly ionized helium
(d) Doubly ionized lithium
digAnsr: d
Ans : (d) Doubly ionized lithium
For a hydrogen-like ion with Z protons in the nucleus, the radius of the nth state is given by
`` {r}_{\,\mathrm{\,n\,}}=\frac{{n}^{2}{a}_{0}}{Z}``
Here, a₀ = 0.53 pm
For lithium,
Z = 3
Therefore, the radius of the first orbit for doubly ionised lithium will be minimum.
Page No 383:

Qstn #6
In which of the following systems will the wavelength corresponding to n = 2 to n = 1 be minimum?
(a) Hydrogen atom
(b) Deuterium atom
(c) Singly ionized helium
(d) Doubly ionized lithium
digAnsr: d
Ans : (d) Doubly ionized lithium
The wavelength corresponding the transition from n₂ to n₁ is given by
`` \frac{1}{\lambda }=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)``
Here,
R = Rydberg constant
Z = Atomic number of the ion
From the given formula, it can be observed that the wavelength is inversely proportional to the square of the atomic number.
Therefore, the wavelength corresponding to n = 2 to n = 1 will be minimum in doubly ionized lithium ion because for lithium, Z = 3.
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Qstn #7
Which of the following curves may represent the speed of the electron in a hydrogen atom as a function of the principal quantum number n?
Figure
Ans : (c)
The speed (v) of electron can be expressed as
`` v=\frac{Z{e}^{2}}{2{\in }_{0}hn}`` ....(1)
Here,
Z = Number of protons in the nucleus
e = Magnitude of charge on electron charge
n = Principal quantum number
h = Planck's constant
It can be observed from equation (1) that the velocity of electron is inversely proportional to the principal quantum number (n).
Therefore, the graph between them must be a rectangular hyperbola.
The correct curve is
(c).
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Qstn #8
As one considers orbits with higher values of n in a hydrogen atom, the electric potential energy of the atom
(a) decreases
(b) increases
(c) remains the same
(d) does not increase
digAnsr: b
Ans : (b) increases
The electric potential energy of hydrogen atom with electron at the nth state is given by
`` V=-\frac{2\times 13.6}{{n}^{2}}``
As the value of n increases, the potential energy of the hydrogen atom also increases, i.e. the atom becomes less bound as n increases.
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Qstn #9
The energy of an atom (or ion) in its ground state is -54.4 eV. It may be
(a) hydrogen
(b) deuterium
(c) He⁺
(d) Li⁺⁺
digAnsr: c
Ans : (c) He⁺
The total energy of a hydrogen-like ion, having Z protons in its nucleus, is given by
`` E=-\frac{13.6{Z}^{2}}{{n}^{2}}`` eV
Here, n = Principal quantum number
For ground state,
n = 1
∴ Total energy, E = `` -`` 13.6 Z² eV
For hydrogen,
Z = 1
∴ Total energy, E = `` -`` 13.6 eV
For deuterium,
Z = 1
∴ Total energy, E = `` -`` 13.6 eV
For He⁺,
Z = 2
∴ Total energy, E = `` -`` 13.6×2² = `` -`` 54.4 eV
For Li⁺⁺,
Z = 3
∴ Total energy, E = `` -`` 13.6×3² = `` -`` 122.4 eV
Hence, the ion having an energy of -54.4 eV in its ground state may be He⁺.
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Qstn #10
The radius of the shortest orbit in a one-electron system is 18 pm. It may be
(a) hydrogen
(b) deuterium
(c) He⁺
(d) Li⁺⁺
digAnsr: d
Ans : (d) Li⁺⁺
The radius of the nth orbit in one electron system is given by
`` {r}_{\,\mathrm{\,n\,}}=\frac{{n}^{2}{a}_{0}}{Z}``
Here, a₀ = 53 pm
For the shortest orbit,
n = 1
For hydrogen,
Z = 1
∴ Radius of the first state of hydrogen atom = 53 pm
For deuterium,
Z= 1
∴ Radius of the first state of deuterium atom = 53 pm
For He⁺,
Z = 2
∴ Radius of He⁺ atom = `` \frac{53}{2}\,\mathrm{\,pm\,}=26.5\,\mathrm{\,pm\,}``
For Li⁺⁺,
Z = 3
∴ Radius of Li⁺⁺ atom = `` \frac{53}{3}\,\mathrm{\,pm\,}=17.66\,\mathrm{\,pm\,}\approx 18\,\mathrm{\,pm\,}``
The given one-electron system having radius of the shortest orbit to be 18 pm may be Li⁺⁺.
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Qstn #11
A hydrogen atom in ground state absorbs 10.2 eV of energy. The orbital angular momentum of the electron is increased by
(a) 1.05 × 10^-34 J s
(b) 2.11 × 10^-34 J s
(c) 3.16 × 10^-34 J s
(d) 4.22 × 10^-34 J s
digAnsr: a
Ans : (a) 1.05 × 10^-34 J s
Let after absorption of energy, the hydrogen atom goes to the nth excited state.
Therefore, the energy absorbed can be written as
`` 10.2=13.6\times \left(\frac{1}{{1}^{2}}-\frac{1}{{n}^{2}}\right)``
`` \Rightarrow \frac{10.2}{13.6}=1-\frac{1}{{n}^{2}}``
`` \Rightarrow \frac{1}{{n}^{2}}=\frac{13.6-10.2}{13.6}``
`` \Rightarrow \frac{1}{{n}^{2}}=\frac{3.4}{13.6}``
`` \Rightarrow {n}^{2}=4``
`` \Rightarrow n=2``
`` ``
The orbital angular momentum of the electron in the nth state is given by
`` {L}_{\,\mathrm{\,n\,}}=\frac{nh}{2\pi }``
Change in the angular momentum, `` ∆L=\frac{2h}{2\pi }-\frac{h}{2\pi }=\frac{h}{2\pi }``
∴ `` ∆L=1.05\times {10}^{-34}\,\mathrm{\,Js\,}``
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Qstn #12
Which of the following parameters are the same for all hydrogen-like atoms and ions in their ground states?
(a) Radius of the orbit
(b) Speed of the electron
(c) Energy of the atom
(d) Orbital angular momentum of the electron
digAnsr: d
Ans : (d) Orbital angular momentum of the electron
According to Bohr's atomic theory, the orbital angular momentum of an electron in a one-electron system is given by
`` {L}_{\,\mathrm{\,n\,}}=\frac{nh}{2\,\mathrm{\,\pi \,}}``
Here,
n = Principal quantum number
The angular momentum is independent of the atomic number of the one-electron system. Therefore, it is same for all hydrogen-like atoms and ions in their ground states.
The other parameters given here are dependent on the atomic number of the hydrogen-like atom or ion taken.
Page No 383:

Qstn #13
In a laser tube, all the photons
(a) have same wavelength
(b) have same energy
(c) move in same direction
(d) move with same speed
digAnsr: d
Ans : (d) move with same speed
All the photons emitted in the laser move with the speed equal to the speed of light (c = 3×10⁸ m/s).
Ideally, the light wave through the laser must be coherent, but in practical laser tubes, there is some deviation from the ideal result. Thus, the photons emitted by the laser have little variations in their wavelengths and energies as well as the directions, but the velocity of all the photons remains same.
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#
Section : iii

Qstn #1
In a laboratory experiment on emission from atomic hydrogen in a discharge tube, only a small number of lines are observed whereas a large number of lines are present in the hydrogen spectrum of a star. This is because in a laboratory
(a) the amount of hydrogen taken is much smaller than that present in the star
(b) the temperature of hydrogen is much smaller than that of the star
(c) the pressure of hydrogen is much smaller than that of the star
(d) the gravitational pull is much smaller than that in the star
digAnsr: b
Ans : (b) the temperature of hydrogen is much smaller than that of the star
The number of lines of the hydrogen spectrum depends on the excitation of the hydrogen atom. This is dependent on the heat energy absorbed by the hydrogen atoms. More the temperature of the hydrogen sample, more is the heat energy. The temperature of hydrogen at the star is much more than that can be produced in the laboratory. Hence, less number of lines are observed in the hydrogen spectrum in the laboratory than that in a star.
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Qstn #2
An electron with kinetic energy 5 eV is incident on a hydrogen atom in its ground state. The collision
(a) must be elastic
(b) may be partially elastic
(c) must be completely inelastic
(d) may be completely inelastic
digAnsr: a
Ans : (a) must be elastic.
The minimum energy required to excite a hydrogen atom from its ground state to 1st excited state is approximately 10 eV. As the incident electron energy is not sufficient for excitation of the hydrogen atom so electron will not get absorbed in the hydrogen atom so it can not be an inelastic collision. Also this collision can not be partially elastic because in an partially elestic collision, there is a net loss on kinetic energy. If the energy is lost then corresponding amount of heat shlould have been produced but it is not so which implies that the collision is completely elastic.
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Qstn #3
Which of the following products in a hydrogen atom are independent of the principal quantum number n? The symbols have their usual meanings.
(a) vn
(b) Er
(c) En
(d) vr
digAnsr: a,b
Ans : (a) vn
(b) Er
Relations for energy, radius of the orbit and its velocity are given by
`` E=-\frac{\,\mathrm{\,m\,}{Z}^{2}{\,\mathrm{\,e\,}}^{4}}{8{{\in }_{0}}^{2}{\,\mathrm{\,h\,}}^{2}{n}^{2}}``
`` r=\frac{{\in }_{0}{\,\mathrm{\,h\,}}^{2}{n}^{2}}{{\,\mathrm{\,\pi mZe\,}}^{2}}``
`` v=\frac{Z{\,\mathrm{\,e\,}}^{2}}{2{\in }_{0}\,\mathrm{\,h\,}n}``
Where
Z : the atomic number of hydrogen like atom
e : electric charge
h : plank constant
m : mass of electron
n : principal quantam number of the electron
`` {\in }_{0}`` : permittivity of vacuum
From these relations, we can see that the products independent of n are vn, Er.
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